Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.4 - Power and Efficiency - Problems - Page 211: 60

Answer

$v=0.285m/s$

Work Step by Step

We can determine the required speed as follows: $P=\frac{U_{1\rightarrow 2}}{t}$ $\implies P=\frac{(80)(9.81)(4)}{t}=100$ This simplifies to: $t=31.4s$ Now $v=\frac{S}{t}$ We plug in the known values to obtain: $v=\frac{\sqrt{(32\times 0.25)^2+(4)^2}}{31.4}$ This simlifies to: $v=0.285m/s$
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